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$y'' - (ax^2 + b) y = 0$ differential equation problem

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0

$begingroup$

I need to figure out when

$$int_{-infty}^{infty}y(x)dx$$

converges

where $y(x)$ is the solution to the differential equation $y'' - (ax^2 + b) y = 0$

I have found out so far that the solution is given in the forms of parabolic cylinder functions which solves differentials on the form $y'' + (ax^2+bx+c)y = 0$

But in order to make use of the parabolic cylinder functions I have to make sure it is on the form

$$y'' - left(dfrac{1}{4}x^2+aright)y = 0$$

How can I rewrite it?

ordinary-differential-equations

share|cite|improve this question

edited Mar 20 at 18:55

John Davis

asked Mar 20 at 18:51

John DavisJohn Davis

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $a$ in your DE is $1/4$ in the parabolic cylinder DE, and $b$ in your DE is $a$ in the parabolic cylinder DE.
  $endgroup$
  – Adrian Keister
  Mar 20 at 18:58
  

  
  
  
  

add a comment | 

0

$begingroup$

I need to figure out when

$$int_{-infty}^{infty}y(x)dx$$

converges

where $y(x)$ is the solution to the differential equation $y'' - (ax^2 + b) y = 0$

I have found out so far that the solution is given in the forms of parabolic cylinder functions which solves differentials on the form $y'' + (ax^2+bx+c)y = 0$

But in order to make use of the parabolic cylinder functions I have to make sure it is on the form

$$y'' - left(dfrac{1}{4}x^2+aright)y = 0$$

How can I rewrite it?

ordinary-differential-equations

share|cite|improve this question

edited Mar 20 at 18:55

John Davis

asked Mar 20 at 18:51

John DavisJohn Davis

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $a$ in your DE is $1/4$ in the parabolic cylinder DE, and $b$ in your DE is $a$ in the parabolic cylinder DE.
  $endgroup$
  – Adrian Keister
  Mar 20 at 18:58
  

  
  
  
  

add a comment | 

$begingroup$

I need to figure out when

$$int_{-infty}^{infty}y(x)dx$$

converges

where $y(x)$ is the solution to the differential equation $y'' - (ax^2 + b) y = 0$

I have found out so far that the solution is given in the forms of parabolic cylinder functions which solves differentials on the form $y'' + (ax^2+bx+c)y = 0$

But in order to make use of the parabolic cylinder functions I have to make sure it is on the form

$$y'' - left(dfrac{1}{4}x^2+aright)y = 0$$

How can I rewrite it?

ordinary-differential-equations

share|cite|improve this question

edited Mar 20 at 18:55

John Davis

asked Mar 20 at 18:51

John DavisJohn Davis

11

$endgroup$

share|cite|improve this question

edited Mar 20 at 18:55

John Davis

asked Mar 20 at 18:51

John DavisJohn Davis

11

share|cite|improve this question

edited Mar 20 at 18:55

John Davis

asked Mar 20 at 18:51

John DavisJohn Davis

11

asked Mar 20 at 18:51

John DavisJohn Davis

11

asked Mar 20 at 18:51

John DavisJohn Davis

11